4% or 40% that is the question

Hi all, I am having an argument about the probability of winning a raffle. Now I am no math wizard but I would like to think I have a grasp for the simple.I may be wrong but we will leave that up to you to decide.

Thanks for the help.

The question is what are the odds or the % chance of winning a raffle if;

There are 10,000 entriesYou have 400 of them and,There are 10 entries drawn one at a time.

Hi all, I am having an argument about the probability of winning a raffle. Now I am no math wizard but I would like to think I have a grasp for the simple.I may be wrong but we will leave that up to you to decide.

Thanks for the help.

The question is what are the odds or the % chance of winning a raffle if;

There are 10,000 entriesYou have 400 of them and,There are 10 entries drawn one at a time.

What is you chance of winning on any one of the 10.

So 400 of 10000 and 10 chance to win.

The probability that the first ticket drawn is not one of yours is .
The prob that the 2nd is not one of yours is
The prob that the 3rd is not one of yours is
:
:
The prob that the 10th is not one of yours is

Rather than ask what is wrong with what you suggest, ask what justification
do you have for averaging these numbers?

That it can't be right is clear from the observation that if only one ticket
was drawn the chance of winning would be 4%, but drawing 10 tickets
increases the chance of winning substantialy. hence what you have done
can't be right and we don't have to bother pointing out exactly where it
goes wrong as:

1. You have no justification for it
2. It gives answers that can't be right

Why is the answer not right?

We have spoke with over 30 people, some who are odds makers in a casino, some are engineers, and we are split on which answer is correct... So again the question is, "Why is your answer right over the other one?" Please someone explain it to us in plain English.

We have spoke with over 30 people, some who are odds makers in a casino, some are engineers, and we are split on which answer is correct... So again the question is, "Why is your answer right over the other one?" Please someone explain it to us in plain English.

1. There is no guarantee that my answer is right, at the verry least you
need to check my arithmetic.

2. It is easier to compute the probability that you dont win, and as the two
events : You win, You don't win are mutualy exclusive:

prob(you win) = 1-prob(you don't win).

(note: here "you win" means one or more of your tickets are winners)

3. The tickets are drawn without replacement, so after N non-winning tickets
are drawn there are 9600-N non-winning tickets left out of 10000-N tickets.

4. Thus there are 9600x9599x..x9591 (equally likely) ways of picking 10 non
winning tickets out of a total of 10000x9999x...x9991 (equally likely) ways of
picking a ticket so the probability that none of the tickets are winners is:

The entries are drawn one at a time. If you lose on the first drawing, then there are now 9999 total entries left, and you still have 400 of them. To not win, you need any of the remaining 9599 entries. This pattern repeats itself nine more times.

Divide by zero, your answer yields an actually lower answer, 33.5167%. Luckily, the sampling size is so big, that your answer is correct up to one decimal place.

The entries are drawn one at a time. If you lose on the first drawing, then there are now 9999 total entries left, and you still have 400 of them. To not win, you need any of the remaining 9599 entries. This pattern repeats itself nine more times.

Divide by zero, your answer yields an actually lower answer, 33.5167%. Luckily, the sampling size is so big, that your answer is correct up to one decimal place.

It's a 2-line Monte-Carlo simulation and 1,000,000 replications gives an
estimate of 33.51%.